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    (Original post by 0x2a)
    I've read The Music of the Primes, although I didn't take everything in. I'm pretty sure the use of modular arithmetic they talked about was for Fermat's Little Theorem and its use in RSA probably. You might want to check Fermat's Little Theorem out, it was used on this thread before.
    Ah yes that sounds familiar, thanks.
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    Here you go ^.^
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    Is reducing it to:

    -2 3
    -3 2

    Right? (I may seem like I'm answering my own question, but these Cambridge STEP questions do not give answers, so I'm basically working it out myself too)
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    (Original post by Smaug123)
    Don't worry - it turns out that the problem is quite hard using GCSE maths. AS and A-level is nothing like this - they'll as good as tell you exactly what to do in each question
    I'll give you a step-by-step answer instead - do you know about partial differentiation? (Or any differentiation at all - we can work with that.)
    I did add maths, so I know basic differentiation (x^2 goes to 2x, x goes to 1, 2 disappears). Never heard of partial differentiation
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    (Original post by paradoxicalme)
    I did add maths, so I know basic differentiation (x^2 goes to 2x, x goes to 1, 2 disappears). Never heard of partial differentiation
    I explained partial differentiation in a nutshell at http://www.thestudentroom.co.uk/show...=10&p=43648264 .
    If you don't know \dfrac{d}{dx} \log(x) = \dfrac{1}{x}, then here is the proof:
    Spoiler:
    Show
    \exp(\log(x)) = x, so \dfrac{d}{dx} \exp(\log(x)) = 1. But the LHS is \dfrac{d}{dx}[\log(x)] \exp(\log(x)) by the chain rule, and because \dfrac{d}{dx}\exp(x) = \exp(x). So \dfrac{d}{dx}[\log(x)] = \dfrac{1}{\exp(\log(x))} = \dfrac{1}{x}, equating the LHS with 1.
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    (Original post by Smaug123)
    I explained partial differentiation in a nutshell at http://www.thestudentroom.co.uk/show...=10&p=43648264 .
    If you don't know \dfrac{d}{dx} \log(x) = \dfrac{1}{x}, then here is the proof:
    Spoiler:
    Show
    \exp(\log(x)) = x, so \dfrac{d}{dx} \exp(\log(x)) = 1. But the LHS is \dfrac{d}{dx}[\log(x)] \exp(\log(x)) by the chain rule, and because \dfrac{d}{dx}\exp(x) = \exp(x). So \dfrac{d}{dx}[\log(x)] = \dfrac{1}{\exp(\log(x))} = \dfrac{1}{x}, equating the LHS with 1.
    Silly question but why does the LHS become that by the chain rule?

    I got this \dfrac{d}{dx}(e)^{logx}=logx(e)^  {logx-1} . 0 ? which equals 0 haha
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    (Original post by Robbie242)
    Silly question but why does the LHS become that by the chain rule?

    I got this \dfrac{d}{dx}(e)^{logx}=logx(e)^  {logx-1} . 0 ? which equals 0 haha
    You're not differentiating correctly. \dfrac{d}{dx} f(g(x)) = g'(x) f'(g(x)); substitute g(x) = \log(x), f(x) = \exp(x) and solve for g'(x).
    What you have worked out is \dfrac{d}{de} \exp(\log(x)), which is odd indeed given that e is constant.
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    (Original post by Applequestria)
    Is reducing it to:

    -2 3
    -3 2

    Right? (I may seem like I'm answering my own question, but these Cambridge STEP questions do not give answers, so I'm basically working it out myself too)
    That is incorrect. Can you see why this is the case? Using a quick check we can see that A would never have any reason to choose the bottom row (clearly something is wrong as the question asserts it must be reduced to a 2x2 matrix!) We also notice that there we can place a straight line that separates negative and positive terms.

    Your error was in your 3rd dominance argument.

    The desired 2x2 matrix is:

    4 -5
    -2 3

    This clearly cannot be reduced by using dominance arguments. The most intuitive way to see this is to think of it as an array of arrows based on what numbers are higher than others.

    In this case, we can visualize there as being four arrows whereby each pair points to opposite corners. Such as array necessary for irreducibility as there are four arrows and only four possible directions with which to point. A sufficient condition would also include as array whereby the 'arrows' all point in to the tail of the proceeding one. Alternatively, you could compare 'minimax' and 'maximin' to establish 'Nash equilibrium'.

    We can then go on to solve this as a linear programming problem for A by letting the proportion of the time it plays each row be p and 1-p respectively. We could also use simplex if we wanted (though, in this case, this seems rather unnecessary).

    I won't write the solution up. If you want to I can check it for you?
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    (Original post by Smaug123)
    You're not differentiating correctly. \dfrac{d}{dx} f(g(x)) = g'(x) f'(g(x)); substitute g(x) = \log(x), f(x) = \exp(x) and solve for g'(x).
    What you have worked out is \dfrac{d}{de} \exp(\log(x)), which is odd indeed given that e is constant.
    oh thanks for the clarification
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    (Original post by Smaug123)
    Yeah, sorry - the thread was a bit derailed

    How about:
    (spoiler defines a "group")
    Spoiler:
    Show

    I define a "group" to be a set X, and an operation +, such that the following four axioms are true:
    1) a+b is in X for every a,b in X
    2) there is an element, e (the identity), such that a+e = a = e+a for all a in X
    3) every element a in X has a corresponding element z in X with a+z = z+a = e
    4) (a+b)+c = a+(b+c) for all a,b,c in X.

    So the set "the integers, with +" is a group:
    1) a+b is an integer for all integers a,b
    2) such an element is 0
    3) this is true: set z=-a
    4) this is blindingly obviously true.

    The set "{1,2,3,4} with the operation *" is a group, where in the following table, the entry in the mth row and the nth column gives m*n:
    1234
    2143
    3412
    4321
    (it's clear from the table that the first axiom holds; 1 is our e; each element a has a*a=e; and it can be tediously checked that the fourth axiom holds.)

    Groups do not have to be "commutative" - that is, a+b isn't necessarily b+a. However, the first example is of size 6, and it's a pain (and not particularly helpful) to write out the table.

    The question is:
    Show that the identity is unique in a group. That is, show that if there are two elements e,f such that e+a = a+e = f+a = a+f = a for all a in the set X, then e=f.

    This may take you a while to get your head around, but it's a good exercise in exploring a definition. That skill is very much not taught at school - even at A-level - so don't be disheartened if you don't know what to do. Write things down and see if they help!
    Good luck!
    I see your valiant attempt at introducing us to groups has been ignored.

    For your question at the end is my answer satisfactory?

    e + a = f + a
    e + a + z = f + a + z
    e + e = f + e (Property 3)
    e = f (Property 2)
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    (Original post by 0x2a)
    I see your valiant attempt at introducing us to groups has been ignored.

    For your question at the end is my answer satisfactory?

    e + a = f + a
    e + a + z = f + a + z
    e + e = f + e (Property 3)
    e = f (Property 2)
    Yep, I think so
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    (Original post by Jkn)
    That is incorrect. Can you see why this is the case? Using a quick check we can see that A would never have any reason to choose the bottom row (clearly something is wrong as the question asserts it must be reduced to a 2x2 matrix!) We also notice that there we can place a straight line that separates negative and positive terms.

    Your error was in your 3rd dominance argument.

    The desired 2x2 matrix is:

    4 -5
    -2 3

    This clearly cannot be reduced by using dominance arguments. The most intuitive way to see this is to think of it as an array of arrows based on what numbers are higher than others.

    In this case, we can visualize there as being four arrows whereby each pair points to opposite corners. Such as array necessary for irreducibility as there are four arrows and only four possible directions with which to point. A sufficient condition would also include as array whereby the 'arrows' all point in to the tail of the proceeding one. Alternatively, you could compare 'minimax' and 'maximin' to establish 'Nash equilibrium'.

    We can then go on to solve this as a linear programming problem for A by letting the proportion of the time it plays each row be p and 1-p respectively. We could also use simplex if we wanted (though, in this case, this seems rather unnecessary).

    I won't write the solution up. If you want to I can check it for you?
    Oh, whatever

    I haven't even studied D2 yet xD

    But yeah, I see where I went wrong ^.^ And yeah, I'll try and work out the rest :P
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    (Original post by Smaug123)
    You're not differentiating correctly. \dfrac{d}{dx} f(g(x)) = g'(x) f'(g(x)); substitute g(x) = \log(x), f(x) = \exp(x) and solve for g'(x).
    What you have worked out is \dfrac{d}{de} \exp(\log(x)), which is odd indeed given that e is constant.
    I am not getting this at all I keep getting 1 on both sides and its driving me nuts
    g(x)=logx g'(x) is what we are looking for f(x)=e^{x}
    f(g(x))=e^{logx}=x
    f'(g(x))=1 \dfrac{d}{dx} f(g(x))= 1

    I definitely know I'm doing f'(g(x)) wrong then. Any help? Sorry for being a bit of an noob here
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    (Original post by Robbie242)
    I am not getting this at all I keep getting 1 on both sides and its driving me nuts
    g(x)=logx g'(x) is what we are looking for f(x)=e^{x}
    f(g(x))=e^{logx}=x
    f'(g(x))=1 \dfrac{d}{dx} f(g(x))= 1

    I definitely know I'm doing f'(g(x)) wrong then. Any help? Sorry for being a bit of an noob here
    f'(g(x)) = \exp'(\log(x)) = \exp(\log(x)) = x.
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    (Original post by Smaug123)
    f'(g(x)) = \exp'(\log(x)) = \exp(\log(x)) = x.
    don't worry I understand it now, was being a bit stupid with my exponents! Thanks anyway
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    Question:

    A function f(x) is defined by 2x2 + 4x + k

    Find the values of k such that f(x) has equal roots.
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    (Original post by tohaaaa)
    Question:

    A function f(x) is defined by 2x2 + 4x + k

    Find the values of k such that f(x) has equal roots.
    Woop! An easy one

    2x2 + 4x + k

    To have one root (or equal roots), the discriminant must equal 0.

    b2 - 4ac = 0
    42 - 4(2)k = 0
    16 - 8k = 0
    k = 2

    Therefore the value that k must be for the function to have equal roots is 2.
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    Find the first three terms in ascending powers of x of the binomial expansion of ln(2x+1), stating the range of values of x for which it is valid.
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    (Original post by fayled)
    Find the first three terms in ascending powers of x of the binomial expansion of ln(2x+1).
    I'll do it this way, since you mention binomial expansion:
    Ignoring constants of integration:
    \ln(2x+1) = \displaystyle\int \dfrac{2}{2x+1} \ dx = \displaystyle\int 2(1 -2x + 4x^2 + . . .) \ dx = 2x - 2x^2 +\dfrac{8x^3}{3} + . . .

    EDIT: Valid for |x|<\dfrac{1}{2}
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    (Original post by joostan)
    I'll do it this way, since you mention binomial expansion:
    Ignoring constants of integration:
    \ln(2x+1) = \displaystyle\int \dfrac{2}{2x+1} \ dx = \displaystyle\int 2(1 -2x + 4x^2 + . . .) \ dx = 2x - 2x^2 +\dfrac{8x^3}{3} + . . .
    Correct. What would your alternative approach have been?
 
 
 
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